2017 Bacteria-Viruses STQ MS

2017 Bacteria and Viruses STQ MS 2017 / H2 / ACJC PRELIM / P2 Q8 (BE)1Speciation events have been observed to occur very frequently in bacteria. It was suggested that the high rate of speciation is due to the high level of variation in bacteria.(a)Transformation and conjugation are two processes which increase the level of variation in bacteria. Distinguish these two processes. ConjugationTransformationGenetic material being transferred1.Transfers the F plasmid or R plasmid1.Transfers fragments of the bacterial chromosome;Donor cell2.Donor cells are F+ cells / carries the F plasmid or R plasmid;2.Donor cells may be lysed cells that releases its DNA;Recipient cell3.Recipient cells are F- cells / do not carry the F plasmid or R plasmid;3.Recipient cells must be competent / secrete the competence factor;Physical contact4.Direct physical contact required between two cells through sex pilus;4.No physical contact required between cells / exogenous DNA taken up directly by recipient cell; [3]Bacterial evolution is one of the most dynamic and exciting areas in current biological research. Over the years, a barrier in this field of research is the difficulty in classifying bacterial species. However, in recent times, new analytical tools in molecular biology have offered new insights into the classification of bacterial species.(b) (i) Suggest why scientists had difficulties in the classification of bacterial species. 1.Bacteria reproduce asexually / by binary fission;2.Unable to determine between species according to biological species concept / bacteria are unable to interbreed to produce fertile viable offspring;3.Horizontal gene transfer/transformation/conjugation between relatively distantly related bacteria / different bacterial species (OWTTE);4.Results in high rates of recombination, making it difficult to determine a single species;5.Different species may be morphologically similar;[2](ii)Explain how analytical molecular tools have helped overcome this barrier in research.1. Molecular tools have helped determine genetic sequences of bacteria / compare genetic sequences of bacteria;2. Allowing scientists to classify bacteria according to genetic distance / phylogenetic distance;3.Provides an objective method (to determine genetic distance);4.Data obtained is quantitative;5.Hence more sensitive to differences between species / data could be used for statistical analyses;[2]

(c)The outer layers of the two types of bacteria with peptidoglycan cell walls known as Gram-positive and Gram-negative bacteria are shown in Fig. 8.1 below. Penicillin is an antibiotic that is known to be effective against only one of the two types of bacteria above.With reference to the information given and your own knowledge, deduce which type of bacteria is susceptible to the action of penicillin and explain why.1. Penicillin is effective only against gram-positive bacteria;2. Gram-negative bacteria have an outer membrane (and lipoproteins) (that surrounds the peptidoglygan layer of the cell wall) / penicillin can directly access the peptidoglycan cell wall in gram-positive bacteria;3. Preventing the action of penicillin as penicillin inhibits the crosslinks in peptidoglycan cell wall;[3](d)Explain how antibiotic-resistant bacteria can become increasingly common in a population of bacteria.1.Horizontal gene transfer (transformation, transduction, conjugation) occurs which increases genetic variation in the bacteria;2.Genetic variation exists in the form of antibiotic sensitivity and antibiotic resistance;3.Antibiotics act as selection pressure;4.Non-resistant bacteria are selected against / bacteria which are resistant to antibiotics are selected for / they have a selective advantage;5.Allele coding for antibiotic resistance passed down to subsequent generations of bacterial cells (during binary fission);6.Over many generations, frequency of allele coding for antibiotic resistance increases in the gene pool;[4][Total: 14 m] Gram-negative bacteriaGram-positive bacteriaFig. 8.1

2017 / H2 / AJC PRELIM / P2 Q55Figure 5.1 represents a bacteria DNA and a eukaryotic chromosome in metaphase of mitosis, not drawn to scale. Fig. 5.1(a)State two ways in which the organization of genes found in these two structures differ and suggest one advantage of this to the bacterium. FeatureEukaryoticProkaryotic Gene organizationMonocistronic genesPolycistronic genes / operonsAdvantage to bacteriaSimultaneous expression of closely-related genes organised in an operonAssociation between DNA and histonesAssociation with histones / scaffolding Proteins- allows for increased structuralcomplexity/folding to higher degree ofcondensation e.g. between euchromatin and heterochromatin statesNo association with histones- does not achieve same level of condensationcomplexity as eukaryote[3](b)In 1946, Joshua Lederberg and Edward Tatum proposed that bacterial cells undergo genetic recombination. To test their hypothesis, they experiments using two bacteria strains of Escherichia coli (E.Coli) with different nutritional requirements. Strain A, B and a mixture of both strains were grown on culture plates containing minimal medium that does not contain essential amino acids. The results are shown in Fig. 5.2.Mutant genes () do not code for enzymes that synthesize amino acids. Note that all five amino acids are required for bacterial growth.Bacterial strainsGenes for biosynthesis of amino acidsMutant genes for biosynthesis of amino acidsAthr+ leu+ thi+met bioBmet+ bio+thr leu thi

Fig. 5.2Another researcher, Bernard Davis also worked with the same hypothesis. In his experiment he constructed a U-tube in which the two arms were separated by a fine filter. The pores of the filter were too small to allow bacteria to pass through but large enough to allow easy passage of the fluid medium, any dissolved substances and free DNA. The results are shown in Fig. 5.3. Fig. 5.3(i)Using the results of the two experiments and your understanding of genetic recombination in bacteria, state the genetic recombination that has taken place between Strain A and B. Explain your answer. Conjugation;Second experiment shows transduction and transformation did not take place;Because no colonies grew on minimal medium agar;genetic recombination occurs only when physical contact is possible between 2 strains;In first experiment, bacteria that grew on minimal medium has DNA that encodes for all essential amino acids/ ref. recombinant bacteria has grown on minimal mediumShowing the genes from Strain A have transferred to Strain B / converse through formation of conjugation tube/ ref. to conjugation tube [6](c)In 2016, a pathogenic strain of E.Coli found on unwashed salad caused food poisoning in 151 people in Britain, leaving two of them dead. Using a named example, describe how such pathogens are usually treated. [3]

Antibiotics PenicillinCompetitive inhibitor to transpeptidases, prevent cell wall synthesis;Resulting in bacterial cell lysis;[Total: 12 m]2017 / H2 / CJC PRELIM / P2 Q8 (ETA)3Transpeptidase is a bacterial enzyme that cross-links cell wall peptides during the formation of bacterial cell walls. The antibiotic penicillin inhibits the activity of transpeptidase. Fig. 8.1 shows part of each of the molecular structures of a cell wall peptide and penicillin. Fig. 8.1(a)Comment on the structure of cell wall peptides and penicillin.ANS [L1] (Novel) [1]2] Shapesimilar configuration ring structure1.Both have similar functional groups C=O [Carbonyl] and COOH [carboxylic groups]2.Both have similar shape / configuration (b)Suggest why the penicillin molecule is an effective inhibitor of transpeptidase.ANS [L2] (Adapted from H2 JJC/ 2017 MYE/ Q4bii) [2]1.As the penicillin molecule/competitive inhibitor is structurally similar to the cell wall peptide/actual substrate, the penicillin molecule can enter and bind/competes with the cell wall peptide for binding at the active site of the transpeptidase;; 2.When the pen